Asked by Aust
Consider the reaction
PCl3(ℓ) → PCl3(g)
at 298 K. If ∆H◦
is 32.5 kJ/mol, ∆S
◦
is
93.3 J/K mol, and ∆G
◦
is 4.7 kJ/mol, what
would be the boiling point of PCl3 at one
atmosphere?
PCl3(ℓ) → PCl3(g)
at 298 K. If ∆H◦
is 32.5 kJ/mol, ∆S
◦
is
93.3 J/K mol, and ∆G
◦
is 4.7 kJ/mol, what
would be the boiling point of PCl3 at one
atmosphere?
Answers
Answered by
DrBob222
dGo = dHo - TdSo
Substitute for dH and dS, set dGo = 0, and solve for T (in kelvin).
dGo = 0 because that is that point that the liquid phase/gas phase is in equilibrium and dGop = 0.
The number you obtain is based on dHo and dSo not changing very much from the value you get from the tables at 298 K.
Substitute for dH and dS, set dGo = 0, and solve for T (in kelvin).
dGo = 0 because that is that point that the liquid phase/gas phase is in equilibrium and dGop = 0.
The number you obtain is based on dHo and dSo not changing very much from the value you get from the tables at 298 K.
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